This is probably a dumb question... here goes. Eigen can solve A x = b, is there also an easy way to solve x A = b?

Really, I'm trying to calculate the gain for a Kalman filter, which is typically written as K = P H (H P H^T + R) ^-1. I'm trying to avoid doing a matrix inverse by solving for the kalman gain, K. Any suggestions?

You can transpose your problem:

x.transpose() = A.transpose().lu().solve(b.transpose());

Note that .transpose() returns an expression, so no copy/transposition overhead.

This is just what I was looking for. I forgot (A B)^T = B^T A^T, thanks!

Also, it's good to know there is no copy/transposition overhead. Eigen is very useful.

Just for the record, in the future we plan to support a more natural syntax like:

x = b * A.lu().inverse();

(that will trigger a call to solve, no explicit inverse computation!)